International Journal Of Computational Engineering Research (ijceronline.com) Vol. 3 Issue. 2
On Presence of Interaction In An Unbalanced TwoWay Random Model 1,
F.C. Eze, 2,P.E. Chigbu
1. Department
of Statistics, Nnamdi Azikiwe University, Awka, Anambra State, Nigeria 2.Department of Statistics, University of Nigeria, Nsukka, Nigeria
Abstract In an unbalanced twoway random model, there is no obvious denominator for testing for the main effects as a result of the presence of interaction. The interaction was removed from the model/data resulting to a reduced model devoid of interaction. The data were transformed by dividing the entries of each cell of the data by the inverse of the square root of the standard error to remove the interaction.
Keywords: Expected mean squares, fractional degrees of freedom, reduced model. 1. Introduction The presence of interaction in twoway analysis of variance could be a serious problem when testing for the main effects. Chow[3] emphasized that testing the treatment effects for the twoway ANOVA model with interaction, the interaction effect has to be tested first, otherwise the result for testing the treatment cannot be interpreted in a statistical meaningful manner. In ANOVA, a large F value provides evidence against the null hypothesis. However, the interaction test should be examined first. The reason for this is that, there is little point in testing the null hypothesis H A or HB if HAB: no interaction effect is rejected, since the difference between any two levels of a main effect also includes an average interaction effect Cabrera and McDougall[1] argued. Moore et al [5] argued that, there are three hypotheses in a twoway ANOVA, with an F test for each. We can test for significance of the main effect A, the main effects of B and AB interaction. It is generally a good practice to examine the test interaction first, since the presence of a strong interaction may influence the interpretation of the main effects. Muller and Fetterman [6] agreed that model reduction in the presence of non significant interaction may be attractive for unbalanced or incomplete design.
2. Methodology Given the model
i 1, 2,..., a yijk i j ij eijk j 1, 2,..., b k 1, 2,..., n ij
(1)
Where yijk is the kth observation in ijth cell,
is the overall mean effect, ij is the effect of the interaction between factor A and factor B,
eijk is a random error components,
nij is the number of observation per cell; and using the BruteForce Method, the expected mean squares (EMS) of the parameters of Equation (1) can be shown to be EMS A e2 k 2 k1 2 EMS B e2 k 2 k2 2 EMS e2 k3 2 EMSe e2
where
k Issn 22503005(online)
N N 1 N i2 i
(2)
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k1 k
k2
( Ni1 nij2 N 1 nij2 i
j
ij
(3)
a 1 1 N N N 2j j
(4)
b 1 ( N j 1 nij2 N 1 nij2 j
i
ij
(5)
b 1 1 2 k3 ( N N i nij N j 1 nij2 N 1 nij2 i
j
j
i
ij
2 n N N ) / ( a 1)(b 1) 3 ij
1 i
1 j
(6)
ij
where
2 E( yi.. y.. ) 2 2 E( y. j . y.. ) 2
(y
i ..
y... ) 2
i
n ( y. j. y... )2 i
n
( y
ij .
2 E( yij. yi.. y. j. y... ) 2
yi.. y. j. y... )2
ij
e2 E( yijk yij. ) 2
n ( yijk yij. )2 ijk
n
k1 , k2 and k3 are the coefficient of the variance components of the interaction for factor A, factor B and the interaction between factor A and factor B respectively. The expected mean squares (EMS) of Equation (1) are presented in the ANOVA Table shown in Table 1. S.V Factor A
d.f a1
SS SS
MS MS
EMS e2 k 2 k1 2
Factor B
b1
SS
MS
e2 k 2 k2 2
AxB
(a1)(b1)
SS
MS
e2 k3 2
Error
Npq
SS e
MSe
e2
Total
N1
SST Table 1: ANOVA Table for Unbalanced data.
From the expected mean squares in Table 1, we can see that the appropriate statistic for testing the no interaction hypothesis H 0 : 0 is 2
Fc
MS MS e
This is because, under H 0 both numerator and denominator of Fc have expectation e . 2
The case is different when testing for H 0 : 0 because the numerator expectation is 2
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International Journal Of Computational Engineering Research (ijceronline.com) Vol. 3 Issue. 2
expectation in Table 1 that is
e2 k1 2 under H 0 .
The case is also the same when testing for
H0 : 2 0
The cause of the above problem is the presence of interaction and when the interaction is removed from the model, the ANOVA Table 1 reduces to the ANOVA Table shown in Table 2. S.V Factor A
d.f a1
SS SS
MS MS
EMS e2 k 2 k1 2
Factor B
b1
SS
MS
e2 k 2 k2 2
Error
Nab+1
SS e
MSe
e2
Total
N1
SST
Table 2: Reduced ANOVA Table for Unbalanced data. From the ANOVA Table 2, it is obvious that the common denominator for testing for the main effects is MSe.
3
Method of Removing the Interaction
Now to remove the interaction from the model we proceed as follows: The least square estimate for the interaction is:
ij yij . yi.. y. j . y... yij . , yi.. , y. j . , y... are the cell means, row means for factor A, column means for factor B and the overall means respectively for the full model. Removing the interaction from Equation (1) we have * yijk yijk yij . yi.. y. j . y...
(7)
i j zijk Dividing Equation (7) by
k and simplifying we have yij*. yi.. y. j. y...
Similarly
yi*.. yi.. and
y...* y... * * * * zijk is the error associated with yijk for the reduced model while yij. , yi.. and y... are the cell means, row means for
factor A and the overall means for the reduced model. It then follows that yi*.. y...* yi .. y... y.*j . y...* y. j . y...
Now
zijk eijk eij . ei .. e. j . e... Issn 22503005(online)
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To transform and normalize the data, we divide the original data by the inverse of the standard error. To do this we need the var( zijk ) . From Equation (1) nij is the number of observations per cell. It then follows that
n
N i.
n
N. j
n
N..
ij
j
ij
i
ij
ij
It can be shown that
E ei..
e2
E e. j .
e2
E eij .
e2
E e...
e2
Ni. N. j nij
N..
Therefore 2 2 2 2 2nij e 3 e2 Variance of zijk e2 e e e nij N i . N. j N i . N. j N..
2nij 1 1 1 3 e2 1 n N N N N N ij i. .j i. .j ..
Therefore
E MSz kij E MSe kij e2 Where 2nij 1 1 1 3 kij 1 n N N N N N ij i. .j i. .j ..
(8)
Therefore
SS z ijk
2 2 1 yijk yij . w jk wij . kij
Where
wijk
yijk kij
wijk is now the transformed data. 4
Illustrative Example
The data below correspond to an experiment in which four different methods for growing crops were tested on four different types of fields (same soil but different light exposure). The yield was measured after the harvest. Because the 3rd method was not tested on the 4th type of field (because of a lack of seeds), and the 2nd method on the 4th type of Issn 22503005(online)
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field (because of a hail storm), the experiment is a typical example of an unbalanced ANOVA. Type of field Method 1 2 3
1 20, 7 35, 52 62, 44
2 39, 17 30, 28 82, 81
3 34, 13 58, 73 69, 84
4 13, 5 64 
Table 3: Source: Kovach Computing Services [4] The model is the same as in Equation (1), where yijk the yield after harvest.
is a constant. i is the mean effects of the method representing factor A. j is the mean effects of the type of field representing factor B. ij is the interaction between the method and type of field. eijk is the error associated with yijk . The sums of squares for the unbalanced data which are analogous to balanced designs are:
SS A Ni yi.. y...
2
(9)
i
SS B N j y. j . y...
2
(10)
j
SS
n y ij
ij .
yi .. y. j . y...
2
(11)
ij
SSe ( yijk yij . )2
(12)
ijk
Using Equations (9), (10), (11) and (12), the sums of squares divided by their respective degrees of freedom for factor A, factor B, the interaction between factor A and factor B and the error term are respectively:MSA = 4749.22, MSB = 641.21, MS =458.7 and MSe = 123. Testing for the main effects will need a special F test because of the unbalanced nature of the design. However, if interaction is non significant or absent, the common denominator for testing for the main effects is the mean square error. Testing for the interaction we have:
F From
MS 458.7 3.73 MSe 123
0.05 we have 3.37 and since 3.73 > 3.37, this shows the presence of interaction in the data, hence the need for F6,9
appropriate transformation of data. Using equation 8 the transformed data are shown in Table 4. Type of field Method
1
2
3
4
1
17.2, 6.02
33.5,
14.6
29.2, 11.2
2
30.5,
45.2
26.1,
24.4
50.5,
63.5
41.6
3
55.2,
39.2
73, 72.1
61.4,
74.8

9.1, 3.5
Table 4: Transformed unbalanced data Having transformed the data, the new sums of squares for the main effects and error are:
SS A Ni yi.. y...
2
(13)
i
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SS B N j y. j . y...
2
(14)
j
SSe ( yijk yij. )2
(15)
ijk
The sums of squares and the corresponding mean sum of squares for the main effects and the error terms has been calculated using Equations (13), (14) and (15) and presented in the ANOVA Table 5 S.V Method Type of field Error Total
d.f 2 3 15 20
SS 7705.35 2066.43 831.01 28602.79
MS 3852.68 688.81 92.33
F ratio 69.54 12.43
Table 5: ANOVA Table From
5
0.05 2,15 =
F
3:68 and
0.05 = 3:29, the main effects are significant. F3,15
Summary and Conclusion
We have shown that in an unbalanced twoway random model, there are no obvious denominator for testing for the main effects which is as a result of the presence of interactions. When interaction is present in an unbalanced twoway random model, we need to construct an appropriate F test for the main effects. To avoid such situation, the interaction from the data/model should be removed to have a valid result when testing for the main effects.
References [1] [2] [3] [4] [5] [6]
[Cabrera and McDougall 2002] Cabrera, J. and McDougall, A. (2002). Statistical Consulting. Springer. [Castrup 2010] Castrup, H. (2010). A WelchSatterthwaite relation for correlated errors.Integrated sciences crop Bakers eld CA 93306. [2003] Chow, S.C. (2003) Encyclopedia of Biopharmaceutical Statistics. p 1027. Informa Health Care. [2011] Kovach Computing Services (2011). "How can I use XLSTAT to run a twoway unbalanced ANOVA with interaction." Anglesey, Wales. [2004] Moore, D.S., McCablee, G., Duckworth, W., and Sclove, S. (2004). Practice of business Statistics, part iv pp 1517. Palgrave Macmillan. [Muller and Fetterman 2002] Muller, K., and Bethel, F. (2002). Regression and Analysis of Variance. An integrated approach using SAS P.381. SAS publishing.
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